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```
i am trying to sum the series
dt=67.2
x1=4;
x2=12;
time=0:dt:dt*10;
fdtime=time(2:11);
n=0:4;
h4=zeros(1,11);
h12=zeros(1,11)
for j=1:11  %to compute solution for 11 times
h4(j) =5*x1+  63.6620*sum(ones(size(n))./(sin(n*pi*x1/10).*exp(-0.01175 *((n).^2).*time(j)))) -38.1972*sum(ones(size(n))./(sin((2*(n)-1/20).^2)*pi*x1.*exp(-0.00294 *((2*n-1).^2)*time(j))))

h12(j) =5*x2+  63.6620*sum(ones(size(n))./(sin(n*pi*x2/10).*exp(-0.01175 *((n).^2).*time(j)))) -38.1972*sum(ones(size(n))./(sin((2*(n)-1/20).^2)*pi*x2.*exp(-0.00294 *((2*n-1).^2)*time(j))))
it gives me sum equal to infinity
the second method i used is as follows by taking sum simple  way puting h4 and h 12 as follows in the above statements
h12(j)=  5*x2+(200/pi) * sum((exp(-0.01175 *((n).^2).*dt)) *sin(n*pi*x2/10)*sin(n*pi*x2/10))-(( 120/pi)*sum((exp(-0.00294 *((2*n-1)^2).*dt))*(sin((2*(n)-1/20)^2))*pi*x2));

h4(j)=  5*x1+(200/pi) * sum((exp(-0.01175 *((n).^2).*dt)) *sin(n*pi*x1/10)*sin(n*pi*x1/10))-(( 120/pi)*sum((exp(-0.00294 *((2*n-1)^2).*dt))*(sin((2*(n)-1/20)^2))*pi*x1));
the answers given in the book are different
Time 	X=4	X=12
(sec)
0	     2	2
134.4	1.078	3.191
268.8	1.108	4.272
403.2	1.340	4.873
537.6	1.543	5.248
infinity	2	6

can someone can, very kindly spare some time to have a look and help me what mistake i have done
take care
```
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12/7/2013 5:47:27 AM
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